Question

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/ h and 40 km/h respectively. The velocity of the car midway between P and Q is

• A. $33.3\, km/h$
• B. $20\sqrt 2\, km/h$
• C. $25\sqrt 2\, km/h$
• D. $35\, km/h$

Answer 1

Answer: (D)

$35\, km/h$

Answer 2

To solve this problem, we can use the equations of motion. We are given that a car is moving with uniform accelerationand passes through two points, P and Q, with velocities of 30 km/h and 40 km/h, respectively. We want to find the velocity of the car at the midpoint between P and Q.

Let's break down the problem step by step:

1. Given data:
- Initial velocity at point P (u) = 30 km/h
- Final velocityat point Q (v) = 40 km/h
- Distance between P and Q (s)

2. We can use the following equation of motion to relate velocity, initial velocity, acceleration, and distance:

$v^2 = u^2 + 2as$

3. We want to find the velocity of the car (V) at the midpoint between P and Q. Let's call this point M.

4. First, let's find the acceleration (a) of the car using the data for points P and Q:

$40^2 = 30^2 + 2a s$

Solving for 'a':

$a = \frac{40^2 - 30^2}{2s} = \frac{1600 - 900}{2s} = \frac{700}{2s} = \frac{350}{s}$

5. Now, we can find the velocity at point M using the same equation of motion:

V^2 = u^2 + 2a \left(\frac{s}{2}\right)$$$$V^2 = (30^2) + 2 \left(\frac{350}{s}\right) \left(\frac{s}{2}\right)$$$$V^2 = 900 + 350$$$$V^2 = 1250

Taking the square root of both sides:

$V = \sqrt{1250} \approx 35.35 \text{ km/h}$

So, the velocity of the car at the midpoint between points P and Q is approximately 35.35 km/h.